Question

Two identical conducting spheres with negligible volume have 2.1 nC and –0.1 nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is ________ $\times$ 10⁻⁹ N. [Given : $4\pi\epsilon_0 = \frac{1}{9 \times 10^9}$ SI unit]

Hint:

For identical conducting spheres, when they touch, the final charge on each is simply the average of their initial charges: $q_{final} = (q_1 + q_2)/2$. This is a direct application of conservation of charge and potential equalization.

Answer 1

Correct Answer: 36

When the two identical conducting spheres are brought into contact, the total charge will be redistributed equally between them.
Total charge $Q_{total} = q_1 + q_2 = (2.1 \text{ nC}) + (-0.1 \text{ nC}) = 2.0$ nC.
Since the spheres are identical, the final charge on each sphere, $q'$, will be half of the total charge:
$q' = \frac{Q_{total}}{2} = \frac{2.0 \text{ nC}}{2} = 1.0$ nC.
So, the charge on each sphere after separation is $q' = 1.0 \times 10^{-9}$ C.
The spheres are then separated by a distance $r = 0.5$ m.
The electrostatic force (F) between them is given by Coulomb's law:
$F = k \frac{q' \cdot q'}{r^2}$, where $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ N·m²/C².
$F = (9 \times 10^9) \frac{(1.0 \times 10^{-9})^2}{(0.5)^2}$.
$F = (9 \times 10^9) \frac{1.0 \times 10^{-18}}{0.25}$.
$F = \frac{9 \times 10^{-9}}{0.25} = 36 \times 10^{-9}$ N.
The question asks for the value of the force in units of $10^{-9}$ N.
The value is 36.